Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $n = \dfrac{2k - 8}{k^2 - 12k + 32} \div \dfrac{k + 10}{k - 8} $
Answer: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{2k - 8}{k^2 - 12k + 32} \times \dfrac{k - 8}{k + 10} $ First factor the quadratic. $n = \dfrac{2k - 8}{(k - 8)(k - 4)} \times \dfrac{k - 8}{k + 10} $ Then factor out any other terms. $n = \dfrac{2(k - 4)}{(k - 8)(k - 4)} \times \dfrac{k - 8}{k + 10} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ 2(k - 4) \times (k - 8) } { (k - 8)(k - 4) \times (k + 10) } $ $n = \dfrac{ 2(k - 4)(k - 8)}{ (k - 8)(k - 4)(k + 10)} $ Notice that $(k - 4)$ and $(k - 8)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ 2(k - 4)\cancel{(k - 8)}}{ \cancel{(k - 8)}(k - 4)(k + 10)} $ We are dividing by $k - 8$ , so $k - 8 \neq 0$ Therefore, $k \neq 8$ $n = \dfrac{ 2\cancel{(k - 4)}\cancel{(k - 8)}}{ \cancel{(k - 8)}\cancel{(k - 4)}(k + 10)} $ We are dividing by $k - 4$ , so $k - 4 \neq 0$ Therefore, $k \neq 4$ $n = \dfrac{2}{k + 10} ; \space k \neq 8 ; \space k \neq 4 $